Induction Proof: $\sum_{k=1}^n k^2$

Question

Prove by induction, the following: $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}6$$ So this is what I have so far:

We will prove the base case for $n=1$: $$\sum_{k=1}^1 1^2 = \frac{1(1+1)(2(1)+1)}6$$ We can see this is true because $1=1$.

Using induction we can assume the statement is true for $n$, we want to prove the statement holds for the case $n+1$: \begin{align*} \sum_{k=1}^{n+1} k^2 :& =\sum_{k=1}^n k^2 + (n+1)^2\\ & = \frac{n(n+1)(2n+1)}6 + (n+1)^2 && \text{by I.H}\\ & = \frac{(n+6)(2n+1)(n+1)^3}6 && \text{algebra} \end{align*}

This is as far as I have gotten in the proof I know that I am almost there, but I am wondering if I missed a step or did the algebra wrong. I know that what I'm supposed to get is $\frac{(n+1)(n+2)(2n+3)}6$ or is that incorrect too? This should be fairly simple, but for some reason it just isn't working for me.

Your "algebra" step is wrong. There's no way to turn the sum of a cubic polynomial and a quadratic polynomial into a fifth-degree polynomial. It looks like you multiplied them, almost. — Thomas Andrews, Jun 19 '15 at 03:26
BTW you can find several posts about this proof on this site. For example: induction proof: $\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$ — Martin Sleziak, Jun 19 '15 at 11:11

Thomas Andrews · Accepted Answer · 2015-06-19T04:03:14.577

2

Without seeing details of your "algebra" step, it is hard to be more specific. Start with: $$\begin{align} \frac{n(n+1)(2n+1)}{6}+(n+1)^2 &= \frac{n(n+1)(2n+1)}{6} + \frac{6(n+1)^2}{6} \\ &= \frac{n+1}{6}(n(2n+1) +6(n+1)) \end{align}$$

edited Jun 19 '15 at 04:03

answered Jun 19 '15 at 03:31

Thomas Andrews

177,126

Thanks for that step. I think my brain just gave up, it's always the simple stuff lol. – SDScott Jun 19 '15 at 03:42

score 1 · Answer 2 · 2015-06-19T09:48:59.163

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$$\displaystyle\sum_{k=1}^{n+1} k^2 :=\sum_{k=1}^n{k^2} + (n+1)^2$$ $$=\frac{n(n+1)(2n+1)}6 + (n+1)^2 \quad\text{by Induction hypothesis}$$ $$=(n+1)\Big(\frac{n(2n+1)}{6}+(n+1)\Big)$$ $$=(n+1)\Big(\frac{2n^2+n+6n+6}{6}\Big)$$ $$=(n+1)\Big(\frac{2n^2+7n+6}{6}\Big)$$ $$=\frac{(n+1)(n+2)(2n+3)}{6}$$

edited Jun 19 '15 at 09:48

answered Jun 19 '15 at 03:52

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Your factorization of $2n^2 + 7n + 6$ is incorrect. To split the linear term, you need two numbers with product $2 \cdot 6 = 12$ and sum $7$. – N. F. Taussig Jun 19 '15 at 09:47
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Yeah ! a small typo was there – Jun 19 '15 at 09:49

Hypergeometricx · Answer 3 · 2015-06-19T17:10:43.957

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This is not an induction proof but might be useful by way of enrichment. $$\sum_{k=1}^n k^2=\frac14\sum_{k=1}^n (2k)^2=\frac14\sum_{k=1}^n \binom {2k}2+\binom {2k+1}2=\frac14\sum_{j=1}^{2n}\binom {j+1}2\\\\ =\frac 14 \binom {2n+2}3=\frac 16n(n+1)(2n+1)\qquad\blacksquare$$

edited Jun 19 '15 at 17:10

answered Jun 19 '15 at 16:49

Hypergeometricx

22,657

Induction Proof: $\sum_{k=1}^n k^2$

3 Answers3